Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 0 0 ≤ x < 1 1.a)Solve: 2 ≤ x + 2 < 3 0 ≤ x < 3 − 2 3 1 3 0 ≤ x < 1 1 3 1 3 3 3 Subtract Multiply b)Graph: © 2007-09 by S-Squared,

Similar presentations


Presentation on theme: "1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 0 0 ≤ x < 1 1.a)Solve: 2 ≤ x + 2 < 3 0 ≤ x < 3 − 2 3 1 3 0 ≤ x < 1 1 3 1 3 3 3 Subtract Multiply b)Graph: © 2007-09 by S-Squared,"— Presentation transcript:

1

2 1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 0 0 ≤ x < 1 1.a)Solve: 2 ≤ x + 2 < 3 0 ≤ x < 3 − 2 3 1 3 0 ≤ x < 1 1 3 1 3 3 3 Subtract Multiply b)Graph: © 2007-09 by S-Squared, Inc. All Rights Reserved.

3 2.The sum of 3 consecutive integers is 66. a)Write a variable equation for the given information. Assign variables: 3x = 63 − 3 x = 21 3 3 b) Find the integers. x + x + 1 + x + 2 = 66 The first integer is x Let, x = the 1 st integer x + 1 = the 2 nd integer 3x + 3 = 66 1 st integer: 21 The second integer is x + 1 2 nd integer: 22 x + 2 = the 3 rd integer The third integer is x + 2 3 rd integer: 23 Subtract Divide Simplify

4 3.The perimeter(P = 2l + 2w) of a rectangle is 24 cm. The length is twice the width. Find the area (A = lw). Assign variables: Let, l = 2w w = w 24 = 2(2w) + 2(w) Substitute into the perimeter formula. Simplify P = 24 24 = 4w + 2w 24 = 6w 6 6 Divide 4 = w Find the length. The length is twice the width. 8 = l Find the area (A = lw). A = (8)(4) Multiply A = 32 Area = 32 cm 2

5 4.On the grid below, draw a coordinate plane and label the following: a)x – axis & y – axis b)Quadrants c)Origin d)Plot the following ordered pairs: a) x – axis & y – axis b) Quadrants c) Origin d) Plot the following ordered pairs: A( − 3, − 5) B( 1, − 3) C( 0, 3) x – axis y – axis Quad I Quad II Quad III Quad IV Coordinates: (0, 0)

6 5.Identify the coordinates of the given points on the graph below and name the quadrant or axis where they lie. CoordinatesQuadrant/ Axis A B C Quad II ( − 4, 3) Quad I ( 2, 1) x- axis ( 5, 0)

7 y = − 4 − 4 − 3 − 3 y = – 4 1 5 (−5) xy − 5 0 5 5 y = – 4 1 5 (0) (0) 1 5 (5) (5) (1) (1) (−1) (−1) (0) (0) 6.a)Create a table to find 3 ordered pairs that are solutions for the following linear equation: y = x – 4 1 5 b)Graph: − 5 − 5 − 4 − 3 y = − 5 − 5 0 Plot the following ordered pairs: (− 5, − 5) (0, − 4) (5, − 3)

8 y = − 2 ─ 4 (−1) y = − 2 ─ 4 (0) y = − 2 ─ 4 (1) y = ─ 4 (2) (0)y = ─ 4 (− 2) y = − 4 y = − 6 y = − 2 xy − 1 0 1 7.Using the following equation: 4x + 2y = − 8 a)Solve for y. 4x + 2y = − 8 2y = − 4x ─ 8 − 4x y = − 2x ─ 4 2 2 2 b)Find 3 ordered pairs that are solutions. y = − 2x ─ 4 − 2 − 4 − 6 − 1 0 1 ( − 1, − 2) ( 0, − 4) (1, − 6) Subtract Divide

9 7.Using the following equation: 4x + 2y = − 8 c)Graph: ( − 1, − 2) ( 0, − 4) (1, − 6) Ordered pairs:


Download ppt "1 2 3 4 5 6 −6 −5 −4 −3 −2 −1 0 0 ≤ x < 1 1.a)Solve: 2 ≤ x + 2 < 3 0 ≤ x < 3 − 2 3 1 3 0 ≤ x < 1 1 3 1 3 3 3 Subtract Multiply b)Graph: © 2007-09 by S-Squared,"

Similar presentations


Ads by Google